InterviewSolution
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InterviewSolution
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`int_(0)^(1)cot^(-1)(1-x +x^(2))dx` का मान ज्ञात कीजिए । |
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Answer» माना `I=int_(0)^(1)cot^(-1)(1-x+x^(2))dx` `=int_(0)^(1)tan^(-1)((1)/(1-x+x^(2)))dx` `=int_(0)^(1)tan^(-1){(x+(1-x))/(1-x+x^(2))}dx` `=int_(0)^(1){tan^(-1)x+tan^(-1_(1-x)}dx` `=int_(0)^(1)tan^(-1)xdx+int_(0)^(1)tan^(-1)(1-x)dx` `=int_(0)^(1)tan^(-1)xdx+int_(0)^(1)tan^(-1){1-(1-x)}dx` `" "` क्योंकि `int_(0)^(a)f(x)d=int_(0)^(a)f(a-x)dx` `=int_(0)^(1)tan^(-1)xdx+int_(0)^(1)tan^(-1)xdx` `=2int_(0)^(1)tan^(-1)xdx=2int_(0)^(1)(tan^(-1)x.1)dx` `=2[(tan^(-1)x)x-int_(0)^(1)(1)/((1+x^(2))).xdx]_(0)^(1)` `=2[(tan^(-1)x)x]_(0)^(1)-2int_(0)^(1)(x)/((1+x^(2)))dx` `=2[(tan^(-1)1).1-0]-[log(1+x^(2))]_(0)^(1)` `=(2xx(pi)/(4))-(log2-log1)=((pi)/(2)-log2)` |
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